Question

PLEASE HURRY IM TIMED!!!



The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as L=10log l/l0, wherel0 =10^-12 and is the least intense sound a human ear can hear. Brandon is trying to take a nap, and he can barely hear his neighbor mowing the lawn. The sound intensity level that Brandon can hear is 10-10. Ahmad, Brandon’s neighbor that lives across the street, is mowing the lawn, and the sound intensity level of the mower is 10-4. How does Brandon’s sound intensity level compare to Ahmad’s?

Answer 1

Answer-

A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.

Solution-

Given that, loudness measured in dB is

$L=10\log \frac{I}{I_0}$

Where,

I   = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's  = 10⁻⁴

Then,

$L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20$

$L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80$

$\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}$

Answer 2

Answer:

a

Step-by-step explanation: